Pedro,
I think you will find that this is the Maximum Reverse Voltage on the
leds (also known as the reverse breakdown voltage), which means that if
there is any more than 5 volts applied across the leads in reverse, the
diode will breakdown and probably start smoking and sputtering and do all
sorts of nasty stuff ...
What you need to look at is the Forward Voltage (denoted as Vf), typical
standard LED's have a Vf of about 2V (between 1.9 & 2.1V). The actual
formula for calculating the resistance based on forward voltage drop, supply
voltage, and maximum continuous forward current is as follows:
R=(E-Vf) x 1000 / I
Where:
R = Resistance needed (Ohms)
E = DC Supply Current (Volts)
I = Total LED current draw (mA)
I've attached a circuit of what I'm pretty sure you are trying to do;
calculated on the values I could work out ... based on a Vf of 2V,
Continuous Forward Current of 30mA, with a supply voltage of 12VDC, which
should be pretty safe ... you can recalculate if need be to accomodate a
higher forward voltage.
Regards,
Jonathan
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----- Original Message -----
From: "Pedro de Oliveira" <olive_@...>
To: <Electronics_101@...>
Sent: Friday, August 03, 2001 12:43 AM
Subject: [Electronics_101] Re: Newbie LED question
Thanks to you all for your excellent suggestions.
The reverse voltage on these LED's is 5V. I take this to mean that
they "want" 5V supply voltage to them.
I have therefore decided (due to your help) to run eight parallel
arrangements of two LED's in series from the 12V supply. My LED's are
30mA and so I would need a 0.5A power supply. I am using a 30A supply
so this should not be a problem (even after distributing power to the
other components).
Since these LED's are 5V, I assume I need a resistor to drop the rest
of the 12V supply (i.e. 2V) so about 66 Ohm (or closest available). I
will connect these to each series array so I will need eight. I am
thinking of connecting the resistors to the negative side of LED.
Does this sound convincing or have I got it all in a mess????
Cheers for your help again.
Pedro de Oliveira
